We rely in the proof on our previous linear analysis, which, in particular, implies \(M\ne \emptyset \)  and that \(\mathcal {T}\)  is well-defined on \(M\) . Let \(q \in M\) . The bound
\(\Vert q\Vert _{L^\infty (L^\infty )}\le \frac{1}{2|k|}\) 
immediately implies that
the non-degeneracy assumption (REF ) holds with \(\underline{\alpha }=\frac{1}{2}\)  and \(\overline{\alpha }=\frac{3}{2}\) .
Moreover, the regularity assumptions of Proposition REF  are satisfied with the uniform bounds
\(\begin{aligned}&\Vert \alpha \Vert _{L^\infty (H^3)}\le |\Omega |^{1/2}+|k| C_{(-\Delta )^{-1}} R, \qquad \Vert \alpha _t\Vert _{L^\infty (H^2)}\le |k| C_{(-\Delta )^{-1}} R, \\&\Vert q_t\Vert _{L^\infty (H^2)}\le C_{(-\Delta )^{-1}} R\,.\end{aligned}\) 
Hence Proposition REF  applied to \(p=\mathcal {T}(q)\) , together with the imposed smallness condition
(REF ) imply that \(\mathcal {T}\)  is well-defined and yield the \(R^2\)  bound in (REF ) on \(\mathcal {T}q=p\) .
To also achieve the \(\frac{1}{2|k|}\)  bound on the \(L^\infty (0,T; L^\infty (\Omega ))\)  norm of \(p\) , we derive a simple energy estimate for the linearization.
Similarly to (REF ), by multiplying equation (REF ) with \(p_t\)  and integrating over space and time, we arrive at
\(\begin{aligned}&\frac{1}{2} \left\lbrace \Vert \sqrt{\alpha (t)}p_t(t)\Vert ^2_{L^2}+c^2\Vert \nabla p(t)\Vert ^2_{L^2} \right\rbrace \Big \vert _0^t+b \int _0^t\Vert \nabla p_t\Vert ^2\, \textup {d}x\textup {d}s\\\lesssim &\, (\Vert \alpha _t\Vert _{L^\infty (L^\infty )}+\Vert q_t\Vert _{L^\infty (L^\infty )})\Vert p_t\Vert ^2_{L^2(L^2)}.\end{aligned}\) 
From here by Gronwall's inequality, we have
\(\begin{aligned}&\Vert p_t(t)\Vert ^2_{L^2}+\Vert \nabla p(t)\Vert ^2_{L^2}\le \, C_1 \exp (C_2RT)(\Vert p_1\Vert ^2_{L^2}+\Vert \nabla p_0\Vert ^2_{L^2})\end{aligned}\) 
for all \(t \in [0,T]\) . Following, e.g., [1]}, the desired \(L^\infty \)  bound on \(p\)  can be obtained by employing the above estimate and Agmon's interpolation inequality [2]}:
\( \Vert p(t)\Vert _{L^\infty } \le C_{\textup {A}} \Vert p(t)\Vert _{H^1}^{1/2}\Vert p(t)\Vert _{H^2}^{1/2}.\) 
Indeed, by also using that \(\Vert p\Vert _{L^\infty (H^2)} \lesssim R\) , we have
\(\begin{aligned}\Vert p\Vert _{L^\infty (L^\infty )}\le C_{\textup {A}}\tilde{C}_1 \exp (\tilde{C}_2RT)(\Vert p_1\Vert ^2_{L^2}+\Vert \nabla p_0\Vert ^2_{L^2})^{1/4}R^{1/2}.\end{aligned}\) 
for some positive constants \(\tilde{C}_1\)  and \(\tilde{C}_2\) , independent of \(b\) . Thus for \(\mathcal {T}(M) \subset M\)  to hold, we need to impose the following condition:
\(C_{\textup {A}}\tilde{C}_1 \exp (\tilde{C}_2RT)(\Vert p_1\Vert ^2_{L^2}+\Vert \nabla p_0\Vert ^2_{L^2})^{1/4}R^{1/2} \le \frac{1}{2|k|},\) 
as claimed.
